
\prob{0000}{中线求长}

\begin{figure}[htbp]
  \centering
  \image{0000}
  \caption{0000：中线求长} \label{fig:0000}
\end{figure}

如图~\ref{fig:0000}，在$\triangle ACD$中，$B$是$AC$的中点，$AD = 3$，$BD = 2$，$CD = 5$，求$AB$的长度。
\problabels{yellow/平面几何, green/长度问题}

\ans{$AB = \sqrt{13}$}

\subsection{倍长中线} \label{subsec:0000-mid}

\begin{figure}[htbp]
  \centering
  \image{0000-mid}
  \caption{\nameref{subsec:0000-mid}：倍长中线，构造3-4-5三角形。} \label{fig:0000-mid}
\end{figure}

基本思路：倍长中线，找出一个3-4-5直角三角形，从而证明$\angle ADB = 90^\circ$。

如图~\ref{fig:0000-mid}，延长$DB$至$D'$，使得$BD = BD'$。连接$AD'$。

\begin{align*}
  &\because   \triangle ABD' \cong \triangle CBD \ \text{（证明省略）} \\
  &\therefore AD' = CD \\
  &\because   CD = 5 \\
  &\therefore AD' = 5 \\
  &\because   BD = BD' \\
  &\therefore DD' = 2BD \\
  &\because   BD = 2 \\
  &\therefore DD' = 4 \\
  &\because   AD = 3 \\
  &\therefore AD^2 + DD'^2 = AD'^2 \\
  &\therefore \angle ADB = 90^\circ \\
  &\therefore AD^2 + BD^2 = AB^2 \\
  &\therefore AB = \sqrt{2^2 + 3^2} = \sqrt{13} \\
\end{align*}

综上，$AB = \sqrt{13}$。
